Question

How would you prepare \(1.0 \mathrm{~L}\) of an aqueous solution of sodium chloride having an osmotic pressure of 15 atm at \(22^{\circ} \mathrm{C}\) ? Assume sodium chloride exists as \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions in solution.

Short answer

To prepare a 1.0 L aqueous solution of sodium chloride with an osmotic pressure of 15 atm at 22°C, first convert the temperature to Kelvin (295.15 K). Then, use the rearranged osmotic pressure formula to calculate the molarity (0.667 mol/L). Next, calculate the amount of sodium chloride needed (approximately 38.97 g) by multiplying the molarity by the volume of the solution. Finally, dissolve the calculated amount of sodium chloride in 1.0 L of water and mix thoroughly to ensure complete dissociation of the sodium chloride into \(\mathrm{Na}^+\) and \(\mathrm{Cl}^-\) ions.

Step-by-step solution

Convert Temperature to Kelvin

First, we need to convert the given temperature from Celsius to Kelvin. To do this, we add 273.15 to the Celsius value: \[T(K) = 22 + 273.15 = 295.15~K\]

Rearrange the Osmotic Pressure Formula for Molarity

Next, we will rearrange the osmotic pressure formula to solve for molarity (M): \[M = \frac{\pi}{RT}\] where π is osmotic pressure, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

Calculate the Molarity of the Solution

Now, we will plug in the given values for osmotic pressure (15 atm) and temperature (295.15 K) into the rearranged formula: \[M = \frac{15~atm}{0.0821 \frac{L\cdot atm}{mol \cdot K} \cdot 295.15~K}\] \[M = 0.667~mol/L\]

Calculate the Amount of Sodium Chloride Needed

Finally, we will calculate the amount of sodium chloride needed to prepare 1.0 L of the solution with the molarity we have found: Amount of Sodium Chloride = Molarity × Volume of the Solution Amount of Sodium Chloride = 0.667 mol/L × 1.0 L Amount of Sodium Chloride = 0.667 mol Now, to convert the moles of sodium chloride to grams, we will use the molar mass of sodium chloride (which is approximately 58.44 g/mol): Amount of Sodium Chloride = 0.667 mol × 58.44 g/mol Amount of Sodium Chloride ≈ 38.97 g

Prepare the Aqueous Solution

To prepare a 1.0 L aqueous solution of sodium chloride with an osmotic pressure of 15 atm at 22°C, you will need to dissolve approximately 38.97 g of sodium chloride in 1.0 L of water. Make sure to mix the solution thoroughly to ensure complete dissociation of the sodium chloride into \(\mathrm{Na}^+\) and \(\mathrm{Cl}^-\) ions.

Key concepts

Converting Celsius to Kelvin

Understanding temperature conversion is crucial in chemistry, as many formulas require temperature to be in the Kelvin scale. The Celsius scale is often used in everyday life. However, scientific calculations necessitate the absolute temperature scale, which is Kelvin (K). To convert Celsius to Kelvin, you simply add 273.15 to the Celsius value. This is because 0 degrees Celsius is equivalent to 273.15 K. For instance, if the temperature is given as 22°C, it needs to be converted before being used in calculations such as osmotic pressure, which are sensitive to absolute temperature.

The formula is straightforward: \[T(K) = T(°C) + 273.15\]. With the example given in the exercise, we get \[ T(K) = 22 + 273.15 = 295.15~K \]. It's a simple yet vital step in ensuring accuracy in subsequent calculations.

Molarity Calculation

Molarity, often denoted as 'M', is a unit of concentration that indicates the number of moles of a solute per liter of solution. It's an important concept in chemistry because it helps to standardize reactions and compare the strength of various solutions.

To calculate molarity, you need two key pieces of information: the amount of the solute in moles, and the volume of the solution in liters. The formula used to find molarity is: \[M = \frac{n}{V}\], where 'M' is the molarity, 'n' is the number of moles of the solute, and 'V' is the volume of the solution in liters. In the provided exercise, after the osmotic pressure formula is rearranged for molarity, we calculate the molarity by dividing osmotic pressure by the product of the gas constant and temperature in Kelvin.

Rearranging Formulas

In many scientific and mathematical problems, including those in chemistry, you have to rearrange formulas to isolate and solve for a specific variable. This process involves understanding the relationships between various elements of the equation and using algebraic operations to get the desired variable on one side of the equation.

In the context of our exercise, to find the molarity from osmotic pressure, we rearrange the osmotic pressure formula to isolate 'M'. The original formula \[\text{π} = MRT\] is manipulated to solve for 'M': \[M = \frac{\text{π}}{RT}\].

Such rearrangement is common in exercises that involve finding concentrations, pressures, volumes, etc., and is an essential skill for problem-solving in chemistry.

Sodium Chloride Dissociation

Sodium chloride, commonly known as table salt, is an ionic compound that dissociates in water to form sodium (Na⁺) and chloride (Cl⁻) ions. This dissociation is important in osmotic pressure calculations because it determines the number of particles in solution, which directly affects the pressure.

When calculating the osmotic pressure of an aqueous sodium chloride solution, you must account for the fact that each formula unit of sodium chloride results in two particles in the solution. Thus, a 1 molar (1 M) solution of sodium chloride would actually result in a 2 M solution in terms of osmotic particles (since both Na⁺ and Cl⁻ contribute to osmotic pressure).

This concept is vital in understanding how ionic compounds behave in solution and how it affects properties like osmotic pressure, boiling point elevation, and freezing point depression.